Extrapolated quantum states, void states and a huge novel class of distillable entangled states

Abstract

A nice and interesting property of any pure tensor product state is that each such state has distillable entangled states at an arbitrarily small distance $\epsilon$ in its neighborhood. We say that such nearby states are $\epsilon$-entangled, and we call the tensor product state in that case, a “boundary separable state,” as there is entanglement at any distance from this “boundary.” Here we find a huge class of separable states that also share the property mentioned above—they all have $\epsilon$-entangled states at any small distance in their neighborhood. Furthermore, the entanglement they have is proved to be distillable. We then extend this result to the discordant/classical cut and show that all classical states (correlated and uncorrelated) have discordant states at distance $\epsilon$, and provide a constructive method for finding $\epsilon$-discordant states.

Appendices

Appendix

The Peres entanglement criterion

Here are a few relevant remarks using the notations of the main article.

1.1 Transpose and partial transpose

Given a Hilbert space ${\mathscr {H}}$ and a basis $\big \{|{i} \rangle \big \}$ (we always assume finite dimensional systems), the transpose is defined by linearity on basis operators $|{i_1} \rangle {\langle {i_2}|}$ by ${T}(|{i_1} \rangle {\langle {i_2}|}) = |{i_2} \rangle {\langle {i_1}|}$. It follows that for any linear operator L, ${\langle {i_1}|}{T}(L)|{i_2} \rangle = {\langle {i_2}|}\,L\,|{i_1} \rangle$. If $\rho$ is a state and $\rho = \sum _i \lambda _i |{\varphi _i} \rangle {\langle {\varphi _i}|}$, one can check that ${T}(\rho ) = \sum _i \lambda _i |{{\overline{\varphi }}_i} \rangle {\langle {{\overline{\varphi }}_i}|}$ where $|{{\overline{\varphi }}} \rangle = \sum _i {\overline{a}}_i |{i} \rangle$ if $|{\varphi } \rangle = \sum _i a_i|{i} \rangle$, ${\overline{a}}_i$ being the complex conjugate of $a_i$. It follows that ${T}(\rho )$ is also a state, with same eigenvalues as $\rho$.

Given a compound system described by ${\mathscr {H}}_A\otimes {\mathscr {H}}_B$, the partial transpose with respect to the A system is simply the operator $({T}\otimes {I})$, i.e.,

$$\begin{aligned} ({T}\otimes {I})\Big (|{i_1} \rangle {\langle {i_2}|}\otimes |{j_1} \rangle {\langle {j_2}|}\Big ) = |{i_2} \rangle {\langle {i_1}|}\otimes |{j_1} \rangle {\langle {j_2}|} \end{aligned}$$

on basis elements. It also follows that for any operator L on ${\mathscr {H}}_A\otimes {\mathscr {H}}_B$

$$\begin{aligned} {\langle {i_1j_1}|}({T}\otimes {I})(L)|{i_2j_2} \rangle = {\langle {i_2j_1}|}\,L\,|{i_1j_2} \rangle \end{aligned}$$

1.2 The Peres Criterion

A state $\rho$ of a bipartite system ${\mathscr {H}}_A\otimes {\mathscr {H}}_B$ is said to be separable if it can be written in the form

$$\begin{aligned} \rho = \sum _i p_i\, \rho ^A_i \otimes \rho ^B_i \quad p_i \ge 0,\ \sum _i p_i = 1 \end{aligned}$$

(6)

where the $\rho ^A_i$ (resp. $\rho ^B_i$) are states of ${\mathscr {H}}_A$ (resp. ${\mathscr {H}}_B$); if $\rho$ is not separable, it is said to be entangled. If $\rho$ is given by (6), then

$$\begin{aligned} ({T}\otimes {I})(\rho ) = \sum _i p_i {T}(\rho ^A_i)\otimes \rho ^B_i \end{aligned}$$

and since the ${T}(\rho ^A_i)$ are states, this implies that $({T}\otimes {I})(\rho )$ is itself a state (and separable). This implies in turn that $({T}\otimes {I})(\rho )$ must be positive semi-definite. As a consequence, if $({T}\otimes {I})(\rho )$ is not positive semi-definite, then $\rho$ is not separable, i.e., it is entangled, that is, the statement of the Peres criterion of entanglement (Peres 1996).

1.3 Checking for positivity

An operator P is positive semi-definite if it is Hermitian and if for all pure states $|{\varphi } \rangle$, ${\langle {\varphi }|} P|{\varphi } \rangle \ge 0$ (iff P has no negative eigenvalue). For any state $\rho$ of ${\mathscr {H}}_A\otimes {\mathscr {H}}_B$, $({T}\otimes {I})(\rho )$ is always Hermitian. To prove that it is not positive semi-definite, we need only find a $|{{\varPsi }} \rangle$ such that ${\langle {{\varPsi }}|}({T}\otimes {I})(\rho )|{{\varPsi }} \rangle < 0$. The partial transpose, however, depends on the basis chosen for ${\mathscr {H}}_A$. We now show (using our notations) that whether $({T}\otimes {I})(\rho )$ is positive semi-definite or not does not depend on the choice of that basis. Indeed, let $|{e_i} \rangle$ be any orthonormal basis of ${\mathscr {H}}_A$. Then $\rho$ can always be written (in a unique way) as $\rho = \sum _{ij} |{e_i} \rangle {\langle {e_j}|}\otimes \rho _{ij}$ where the $\rho _{ij}$ are operators of ${\mathscr {H}}_B$. Let ${T}_e$ be the transpose operator in the basis e, i.e., ${T}_e(|{e_i} \rangle {\langle {e_j}|}) = |{e_j} \rangle {\langle {e_i}|}$. Then

$$\begin{aligned} ({T}_e\otimes I)(\rho )&= \sum _{ij} {T}_e(|{e_i} \rangle {\langle {e_j}|})\otimes \rho _{ij} = \sum _{ij} |{e_j} \rangle {\langle {e_i}|}\otimes \rho _{ij}\\ ({T}\otimes {I})(\rho )&= \sum _{ij} T(|{e_i} \rangle {\langle {e_j}|})\otimes \rho _{ij} = \sum _{ij} |{{\overline{e}}_j} \rangle {\langle {{\overline{e}}_i}|}\otimes \rho _{ij} \end{aligned}$$

The $|{{\overline{e}}_i} \rangle$ also form an orthonormal basis of ${\mathscr {H}}_A$. Now let $|{{\varPsi }_e} \rangle = \sum _i |{e_i} \rangle |{\psi _i} \rangle$ be any pure state of ${\mathscr {H}}_A\otimes {\mathscr {H}}_B$. Then $|{{\varPsi }} \rangle = \sum _i |{{\overline{e}}_i} \rangle |{\psi _i} \rangle$ is also a pure state and

$$\begin{aligned}&{\langle {{\varPsi }}|} ({T}\otimes {I})(\rho )|{{\varPsi }} \rangle = {\langle {{\varPsi }_e}|}({T}_e\otimes I)(\rho )|{{\varPsi }_e} \rangle \nonumber \\&\quad = \sum _{ij} {\langle {\psi _j}|}\rho _{ij}|{\psi _i} \rangle \end{aligned}$$

1.4 Proof of Lemma 10

Proof

Let us assume P is positive semi-definite: $P=\sum _i \lambda _i |{\varphi _i} \rangle {\langle {\varphi _i}|}$ with $\lambda _i\ge 0$. If ${\langle {\varphi }|}P|{\varphi } \rangle = 0$, then $\sum _i \lambda _i |\langle {\varphi } | {\varphi _i} \rangle |^2 = 0$ and $\lambda _i\langle {\varphi } | {\varphi _i} \rangle =0$ for all i and thus ${\langle {\varphi }|}P|{\psi } \rangle = \sum _i \lambda _i \langle {\varphi } | {\varphi _i} \rangle \langle {\varphi _i} | {\psi } \rangle = 0$ for all $|{\psi } \rangle$. $\square$

Distillability

Note that the Peres criterion is not a characterization. If the partial transpose of $\rho$ is positive semi-definite, $\rho$ may still be entangled. Furthermore, if a state $\rho _\textit{ppt-ent}$ is entangled and admits a positive partial transpose then it is not distillable (namely, one cannot distill a singlet state out of many copies of $\rho _ ppt-ent$ via local operations and classical communication). Such states are said to have “bound entanglement.” A characterization of distillable states can be found in Horodecki (1997). Here is the lemma as we use it, as stated in Kraus et al. (2002).

Lemma 25

(Kraus et al. 2002; Horodecki 1997) A state $\rho$ of ${\mathscr {H}}_A\otimes {\mathscr {H}}_B$ is distillable if and only if there exists a positive integer N and a state $|{{\varPsi }} \rangle = |{e_1f_1} \rangle + |{e_2f_2} \rangle$ such that

$$\begin{aligned} {\langle {{\varPsi }}|} \,({T}\otimes {I})(\rho ^{\otimes N})\,|{{\varPsi }} \rangle < 0, \end{aligned}$$

where $\{e_1,e_2\}$ (resp. $\{f_1,f_2\}$) are two unnormalized orthogonal vectors of ${\mathscr {H}}_A^{\otimes N}$ (resp. ${\mathscr {H}}_B^{\otimes N}$).

Proof of Proposition 14 using matrices

When the states $|{ij} \rangle$ are put in lexicographic order, the partial transpose $({T}\otimes {I})$ corresponds to transposing blocks in the block matrix, whereas $({I}\otimes {T})$ corresponds to transposing each of the blocks individually. The matrix of Proposition 14 is a $3\times 3$ block matrix, with $3\times 3$ blocks.

We first calculate for both $\rho _0$ and $\rho _1$ the entries (11, 11) and (01, 10) (row 01, column 10 of their matrix). Those are ${\langle {11}|}\rho _0|{11} \rangle =0$, and ${\langle {01}|}\rho _0|{10} \rangle =0$ for $\rho _0$ and ${\langle {11}|}\rho _1|{11} \rangle =0$ and ${\langle {10}|}\rho _1|{10} \rangle =1/2$ for $\rho _1$. Those values were obtained in the main text. The matrices for $\rho _0$ and $\rho _1$ are then the following (useless entries being kept blank).

Then, the $3\times 3$ block matrix is transposed, giving, respectively, for $({T}\otimes {I})(\rho _0)$ and $({T}\otimes {I})(\rho _1)$ the matrices:

The matrix of $({T}\otimes {I})(\rho _\epsilon ) = (1-\epsilon )({T}\otimes {I})(\rho _0) + \epsilon ({T}\otimes {I})(\rho _1)$ is then

We see clearly that the matrix of $({T}\otimes {I})(\rho _\epsilon )$ has a 0 diagonal entry for which there is a nonzero entry on the corresponding row (or corresponding column). That implies that the matrix is not positive semi-definite and consequently that $\rho _\epsilon$ is entangled.

Of course, the blank values in the density operator for $\rho _1$ could take any value without affecting the result; in fact, any density operator $\rho _1$ such that ${\langle {11}|}\rho _1|{11} \rangle = 0$ and ${\langle {01}|}\rho _1|{10} \rangle \ne 0$ could have been used instead to give entangled states that arbitrarily close to $\rho _0$.